Let $f(x)=(x-2)(x-3)^2$. For what value of $x$ does $f$ have a relative maximum ? Choose 1 answer: Choose 1 answer: (Choice A) A $3$ (Choice B) B $\dfrac{5}{2}$ (Choice C) C $\dfrac{7}{3}$ (Choice D) D $2$
Solution: We can find the relative extrema (i.e. minima and maxima) of $f$ by looking for the intervals where its derivative $f'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $f$ is $f'(x)=(x-3)(3x-7)$. $f'(x)=0$ for $x=3,\dfrac{7}{3}$. Since $f'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=\dfrac{7}{3}$ and $x=3$. $f$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $0$ $1$ $2$ $3$ $4$ $5$ $6$ $(-\infty, \frac{7}{3})$ $(\frac{7}{3},3)$ $\frac{7}{3}$ $(3,\infty)$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $(-\infty,\dfrac{7}{3})$ $x=2$ $f'(2)=1>0$ $f$ is increasing $\nearrow$ $(\dfrac{7}{3},3)$ $x=\dfrac{8}{3}$ $f'\left(\dfrac{3}{8}\right)=-\dfrac{1}{3}<0$ $f$ is decreasing $\searrow$ $(3,\infty)$ $x=4$ $f'(4)=5>0$ $f$ is increasing $\nearrow$ Now let's look at the critical points: $x$ Before After Verdict $\dfrac{7}{3}$ $\nearrow$ $\searrow$ Maximum $3$ $\searrow$ $\nearrow$ Minimum Now we can see that $f$ has a relative maximum at $x=\dfrac{7}{3}$.